TITLE: Is there a way to determine the summation of all "even" elements of an infinite geometric sequence?*
QUESTION [6 upvotes]: *Algebraically and not with by adding them for the first 100 or whatever and estimating
This came up when I came across a question in my probability class where I was asked about a game where 2 players take turns to accomplish a task until one succeeds. Each attempt has a probability of $p = x$ for success (I'm using $x$ so it's general, but in my case, I got .35)
This makes a geometric series of $x(1-x)^{(n-1)}$ for $n$ being a positive integer where $n$ equals nth attempt. 
The question asked was what's the probability the first player wins, I wrote it as: $$\lim_{t\to \infty} \sum_{k=0}^{t}x*(x-1)^{2k}$$
Is there any way to solve this algebraically or something else or do I have to rely on computers for this task?
Sorry for this awful formatting I'm asking this on a phone and I can't post images since I'm new and I can't write latex on mobile.

REPLY [6 votes]: Even terms of a geometric progression still form a geometric progression!
So, using the formula $$\sum\limits_{k=0}^{\infty} p^k = \frac{1}{1-p}$$
we ca easily derive
$$\sum\limits_{k=0}^{\infty} p^{2k} = \sum\limits_{k=0}^{\infty} (p^2)^k = \frac{1}{1-p^2}$$
I hope you can go from here.

REPLY [3 votes]: $$\sum_{k=0}^\infty x(x-1)^{2k} = x\cdot\sum_{k=0}^\infty\left((x-1)^2\right)^k$$
is again a geometric series with $q=(x-1)^2$